Question

it has been demonstrated that \\(\sqrt{97}\\) is between 9.84 and 9.85. if we were to continue this process, it would because \\(\sqrt{97}\\) is .

Explanation:

To solve this, we analyze the nature of \(\sqrt{97}\):

Step 1: Recall the definition of irrational numbers

A number is irrational if it cannot be expressed as a fraction \(\frac{p}{q}\) (where \(p, q\) are integers and \(q
eq 0\)) and its decimal expansion is non - repeating and non - terminating.

Step 2: Analyze \(\sqrt{97}\)

We know that \(9^2=81\) and \(10^2 = 100\). Since 97 is not a perfect square (there is no integer \(n\) such that \(n^2=97\)), by the definition of square roots of non - perfect squares, \(\sqrt{97}\) is an irrational number.

Step 3: Determine the behavior of the approximation process

When approximating an irrational number like \(\sqrt{97}\), since its decimal expansion is non - repeating and non - terminating, if we continue the process of approximating it (for example, finding more decimal places), the approximation process would never end (or be an infinite process) because \(\sqrt{97}\) is an irrational number (its decimal does not terminate or repeat).

For the first blank, a suitable answer is "never end" (or similar phrases like "continue infinitely"). For the second blank, the answer is "an irrational number".

So, filling in the blanks:
First blank: never end
Second blank: an irrational number

Answer:

To solve this, we analyze the nature of \(\sqrt{97}\):

Step 1: Recall the definition of irrational numbers

A number is irrational if it cannot be expressed as a fraction \(\frac{p}{q}\) (where \(p, q\) are integers and \(q
eq 0\)) and its decimal expansion is non - repeating and non - terminating.

Step 2: Analyze \(\sqrt{97}\)

We know that \(9^2=81\) and \(10^2 = 100\). Since 97 is not a perfect square (there is no integer \(n\) such that \(n^2=97\)), by the definition of square roots of non - perfect squares, \(\sqrt{97}\) is an irrational number.

Step 3: Determine the behavior of the approximation process

When approximating an irrational number like \(\sqrt{97}\), since its decimal expansion is non - repeating and non - terminating, if we continue the process of approximating it (for example, finding more decimal places), the approximation process would never end (or be an infinite process) because \(\sqrt{97}\) is an irrational number (its decimal does not terminate or repeat).

For the first blank, a suitable answer is "never end" (or similar phrases like "continue infinitely"). For the second blank, the answer is "an irrational number".

So, filling in the blanks:
First blank: never end
Second blank: an irrational number